3.420 \(\int \frac{(a+b x^2)^{9/2}}{x^7} \, dx\)

Optimal. Leaf size=126 \[ -\frac{105}{16} a^{3/2} b^3 \tanh ^{-1}\left (\frac{\sqrt{a+b x^2}}{\sqrt{a}}\right )-\frac{21 b^2 \left (a+b x^2\right )^{5/2}}{16 x^2}+\frac{35}{16} b^3 \left (a+b x^2\right )^{3/2}+\frac{105}{16} a b^3 \sqrt{a+b x^2}-\frac{\left (a+b x^2\right )^{9/2}}{6 x^6}-\frac{3 b \left (a+b x^2\right )^{7/2}}{8 x^4} \]

[Out]

(105*a*b^3*Sqrt[a + b*x^2])/16 + (35*b^3*(a + b*x^2)^(3/2))/16 - (21*b^2*(a + b*x^2)^(5/2))/(16*x^2) - (3*b*(a
+ b*x^2)^(7/2))/(8*x^4) - (a + b*x^2)^(9/2)/(6*x^6) - (105*a^(3/2)*b^3*ArcTanh[Sqrt[a + b*x^2]/Sqrt[a]])/16

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Rubi [A]  time = 0.0763166, antiderivative size = 126, normalized size of antiderivative = 1., number of steps used = 8, number of rules used = 5, integrand size = 15, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.333, Rules used = {266, 47, 50, 63, 208} \[ -\frac{105}{16} a^{3/2} b^3 \tanh ^{-1}\left (\frac{\sqrt{a+b x^2}}{\sqrt{a}}\right )-\frac{21 b^2 \left (a+b x^2\right )^{5/2}}{16 x^2}+\frac{35}{16} b^3 \left (a+b x^2\right )^{3/2}+\frac{105}{16} a b^3 \sqrt{a+b x^2}-\frac{\left (a+b x^2\right )^{9/2}}{6 x^6}-\frac{3 b \left (a+b x^2\right )^{7/2}}{8 x^4} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*x^2)^(9/2)/x^7,x]

[Out]

(105*a*b^3*Sqrt[a + b*x^2])/16 + (35*b^3*(a + b*x^2)^(3/2))/16 - (21*b^2*(a + b*x^2)^(5/2))/(16*x^2) - (3*b*(a
+ b*x^2)^(7/2))/(8*x^4) - (a + b*x^2)^(9/2)/(6*x^6) - (105*a^(3/2)*b^3*ArcTanh[Sqrt[a + b*x^2]/Sqrt[a]])/16

Rule 266

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Dist[1/n, Subst[Int[x^(Simplify[(m + 1)/n] - 1)*(a
+ b*x)^p, x], x, x^n], x] /; FreeQ[{a, b, m, n, p}, x] && IntegerQ[Simplify[(m + 1)/n]]

Rule 47

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((a + b*x)^(m + 1)*(c + d*x)^n)/(b*
(m + 1)), x] - Dist[(d*n)/(b*(m + 1)), Int[(a + b*x)^(m + 1)*(c + d*x)^(n - 1), x], x] /; FreeQ[{a, b, c, d},
x] && NeQ[b*c - a*d, 0] && GtQ[n, 0] && LtQ[m, -1] &&  !(IntegerQ[n] &&  !IntegerQ[m]) &&  !(ILeQ[m + n + 2, 0
] && (FractionQ[m] || GeQ[2*n + m + 1, 0])) && IntLinearQ[a, b, c, d, m, n, x]

Rule 50

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((a + b*x)^(m + 1)*(c + d*x)^n)/(b*
(m + n + 1)), x] + Dist[(n*(b*c - a*d))/(b*(m + n + 1)), Int[(a + b*x)^m*(c + d*x)^(n - 1), x], x] /; FreeQ[{a
, b, c, d}, x] && NeQ[b*c - a*d, 0] && GtQ[n, 0] && NeQ[m + n + 1, 0] &&  !(IGtQ[m, 0] && ( !IntegerQ[n] || (G
tQ[m, 0] && LtQ[m - n, 0]))) &&  !ILtQ[m + n + 2, 0] && IntLinearQ[a, b, c, d, m, n, x]

Rule 63

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[{p = Denominator[m]}, Dist[p/b, Sub
st[Int[x^(p*(m + 1) - 1)*(c - (a*d)/b + (d*x^p)/b)^n, x], x, (a + b*x)^(1/p)], x]] /; FreeQ[{a, b, c, d}, x] &
& NeQ[b*c - a*d, 0] && LtQ[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntLinearQ[a,
b, c, d, m, n, x]

Rule 208

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[-(a/b), 2]*ArcTanh[x/Rt[-(a/b), 2]])/a, x] /; FreeQ[{a,
b}, x] && NegQ[a/b]

Rubi steps

\begin{align*} \int \frac{\left (a+b x^2\right )^{9/2}}{x^7} \, dx &=\frac{1}{2} \operatorname{Subst}\left (\int \frac{(a+b x)^{9/2}}{x^4} \, dx,x,x^2\right )\\ &=-\frac{\left (a+b x^2\right )^{9/2}}{6 x^6}+\frac{1}{4} (3 b) \operatorname{Subst}\left (\int \frac{(a+b x)^{7/2}}{x^3} \, dx,x,x^2\right )\\ &=-\frac{3 b \left (a+b x^2\right )^{7/2}}{8 x^4}-\frac{\left (a+b x^2\right )^{9/2}}{6 x^6}+\frac{1}{16} \left (21 b^2\right ) \operatorname{Subst}\left (\int \frac{(a+b x)^{5/2}}{x^2} \, dx,x,x^2\right )\\ &=-\frac{21 b^2 \left (a+b x^2\right )^{5/2}}{16 x^2}-\frac{3 b \left (a+b x^2\right )^{7/2}}{8 x^4}-\frac{\left (a+b x^2\right )^{9/2}}{6 x^6}+\frac{1}{32} \left (105 b^3\right ) \operatorname{Subst}\left (\int \frac{(a+b x)^{3/2}}{x} \, dx,x,x^2\right )\\ &=\frac{35}{16} b^3 \left (a+b x^2\right )^{3/2}-\frac{21 b^2 \left (a+b x^2\right )^{5/2}}{16 x^2}-\frac{3 b \left (a+b x^2\right )^{7/2}}{8 x^4}-\frac{\left (a+b x^2\right )^{9/2}}{6 x^6}+\frac{1}{32} \left (105 a b^3\right ) \operatorname{Subst}\left (\int \frac{\sqrt{a+b x}}{x} \, dx,x,x^2\right )\\ &=\frac{105}{16} a b^3 \sqrt{a+b x^2}+\frac{35}{16} b^3 \left (a+b x^2\right )^{3/2}-\frac{21 b^2 \left (a+b x^2\right )^{5/2}}{16 x^2}-\frac{3 b \left (a+b x^2\right )^{7/2}}{8 x^4}-\frac{\left (a+b x^2\right )^{9/2}}{6 x^6}+\frac{1}{32} \left (105 a^2 b^3\right ) \operatorname{Subst}\left (\int \frac{1}{x \sqrt{a+b x}} \, dx,x,x^2\right )\\ &=\frac{105}{16} a b^3 \sqrt{a+b x^2}+\frac{35}{16} b^3 \left (a+b x^2\right )^{3/2}-\frac{21 b^2 \left (a+b x^2\right )^{5/2}}{16 x^2}-\frac{3 b \left (a+b x^2\right )^{7/2}}{8 x^4}-\frac{\left (a+b x^2\right )^{9/2}}{6 x^6}+\frac{1}{16} \left (105 a^2 b^2\right ) \operatorname{Subst}\left (\int \frac{1}{-\frac{a}{b}+\frac{x^2}{b}} \, dx,x,\sqrt{a+b x^2}\right )\\ &=\frac{105}{16} a b^3 \sqrt{a+b x^2}+\frac{35}{16} b^3 \left (a+b x^2\right )^{3/2}-\frac{21 b^2 \left (a+b x^2\right )^{5/2}}{16 x^2}-\frac{3 b \left (a+b x^2\right )^{7/2}}{8 x^4}-\frac{\left (a+b x^2\right )^{9/2}}{6 x^6}-\frac{105}{16} a^{3/2} b^3 \tanh ^{-1}\left (\frac{\sqrt{a+b x^2}}{\sqrt{a}}\right )\\ \end{align*}

Mathematica [C]  time = 0.0119343, size = 39, normalized size = 0.31 \[ \frac{b^3 \left (a+b x^2\right )^{11/2} \, _2F_1\left (4,\frac{11}{2};\frac{13}{2};\frac{b x^2}{a}+1\right )}{11 a^4} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + b*x^2)^(9/2)/x^7,x]

[Out]

(b^3*(a + b*x^2)^(11/2)*Hypergeometric2F1[4, 11/2, 13/2, 1 + (b*x^2)/a])/(11*a^4)

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Maple [A]  time = 0.01, size = 168, normalized size = 1.3 \begin{align*} -{\frac{1}{6\,a{x}^{6}} \left ( b{x}^{2}+a \right ) ^{{\frac{11}{2}}}}-{\frac{5\,b}{24\,{a}^{2}{x}^{4}} \left ( b{x}^{2}+a \right ) ^{{\frac{11}{2}}}}-{\frac{35\,{b}^{2}}{48\,{a}^{3}{x}^{2}} \left ( b{x}^{2}+a \right ) ^{{\frac{11}{2}}}}+{\frac{35\,{b}^{3}}{48\,{a}^{3}} \left ( b{x}^{2}+a \right ) ^{{\frac{9}{2}}}}+{\frac{15\,{b}^{3}}{16\,{a}^{2}} \left ( b{x}^{2}+a \right ) ^{{\frac{7}{2}}}}+{\frac{21\,{b}^{3}}{16\,a} \left ( b{x}^{2}+a \right ) ^{{\frac{5}{2}}}}+{\frac{35\,{b}^{3}}{16} \left ( b{x}^{2}+a \right ) ^{{\frac{3}{2}}}}-{\frac{105\,{b}^{3}}{16}{a}^{{\frac{3}{2}}}\ln \left ({\frac{1}{x} \left ( 2\,a+2\,\sqrt{a}\sqrt{b{x}^{2}+a} \right ) } \right ) }+{\frac{105\,a{b}^{3}}{16}\sqrt{b{x}^{2}+a}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((b*x^2+a)^(9/2)/x^7,x)

[Out]

-1/6/a/x^6*(b*x^2+a)^(11/2)-5/24*b/a^2/x^4*(b*x^2+a)^(11/2)-35/48*b^2/a^3/x^2*(b*x^2+a)^(11/2)+35/48*b^3/a^3*(
b*x^2+a)^(9/2)+15/16*b^3/a^2*(b*x^2+a)^(7/2)+21/16*b^3/a*(b*x^2+a)^(5/2)+35/16*b^3*(b*x^2+a)^(3/2)-105/16*b^3*
a^(3/2)*ln((2*a+2*a^(1/2)*(b*x^2+a)^(1/2))/x)+105/16*a*b^3*(b*x^2+a)^(1/2)

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Maxima [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x^2+a)^(9/2)/x^7,x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.62125, size = 451, normalized size = 3.58 \begin{align*} \left [\frac{315 \, a^{\frac{3}{2}} b^{3} x^{6} \log \left (-\frac{b x^{2} - 2 \, \sqrt{b x^{2} + a} \sqrt{a} + 2 \, a}{x^{2}}\right ) + 2 \,{\left (16 \, b^{4} x^{8} + 208 \, a b^{3} x^{6} - 165 \, a^{2} b^{2} x^{4} - 50 \, a^{3} b x^{2} - 8 \, a^{4}\right )} \sqrt{b x^{2} + a}}{96 \, x^{6}}, \frac{315 \, \sqrt{-a} a b^{3} x^{6} \arctan \left (\frac{\sqrt{-a}}{\sqrt{b x^{2} + a}}\right ) +{\left (16 \, b^{4} x^{8} + 208 \, a b^{3} x^{6} - 165 \, a^{2} b^{2} x^{4} - 50 \, a^{3} b x^{2} - 8 \, a^{4}\right )} \sqrt{b x^{2} + a}}{48 \, x^{6}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x^2+a)^(9/2)/x^7,x, algorithm="fricas")

[Out]

[1/96*(315*a^(3/2)*b^3*x^6*log(-(b*x^2 - 2*sqrt(b*x^2 + a)*sqrt(a) + 2*a)/x^2) + 2*(16*b^4*x^8 + 208*a*b^3*x^6
- 165*a^2*b^2*x^4 - 50*a^3*b*x^2 - 8*a^4)*sqrt(b*x^2 + a))/x^6, 1/48*(315*sqrt(-a)*a*b^3*x^6*arctan(sqrt(-a)/
sqrt(b*x^2 + a)) + (16*b^4*x^8 + 208*a*b^3*x^6 - 165*a^2*b^2*x^4 - 50*a^3*b*x^2 - 8*a^4)*sqrt(b*x^2 + a))/x^6]

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Sympy [A]  time = 7.82761, size = 175, normalized size = 1.39 \begin{align*} - \frac{105 a^{\frac{3}{2}} b^{3} \operatorname{asinh}{\left (\frac{\sqrt{a}}{\sqrt{b} x} \right )}}{16} - \frac{a^{5}}{6 \sqrt{b} x^{7} \sqrt{\frac{a}{b x^{2}} + 1}} - \frac{29 a^{4} \sqrt{b}}{24 x^{5} \sqrt{\frac{a}{b x^{2}} + 1}} - \frac{215 a^{3} b^{\frac{3}{2}}}{48 x^{3} \sqrt{\frac{a}{b x^{2}} + 1}} + \frac{43 a^{2} b^{\frac{5}{2}}}{48 x \sqrt{\frac{a}{b x^{2}} + 1}} + \frac{14 a b^{\frac{7}{2}} x}{3 \sqrt{\frac{a}{b x^{2}} + 1}} + \frac{b^{\frac{9}{2}} x^{3}}{3 \sqrt{\frac{a}{b x^{2}} + 1}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x**2+a)**(9/2)/x**7,x)

[Out]

-105*a**(3/2)*b**3*asinh(sqrt(a)/(sqrt(b)*x))/16 - a**5/(6*sqrt(b)*x**7*sqrt(a/(b*x**2) + 1)) - 29*a**4*sqrt(b
)/(24*x**5*sqrt(a/(b*x**2) + 1)) - 215*a**3*b**(3/2)/(48*x**3*sqrt(a/(b*x**2) + 1)) + 43*a**2*b**(5/2)/(48*x*s
qrt(a/(b*x**2) + 1)) + 14*a*b**(7/2)*x/(3*sqrt(a/(b*x**2) + 1)) + b**(9/2)*x**3/(3*sqrt(a/(b*x**2) + 1))

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Giac [A]  time = 1.85332, size = 143, normalized size = 1.13 \begin{align*} \frac{1}{48} \,{\left (\frac{315 \, a^{2} \arctan \left (\frac{\sqrt{b x^{2} + a}}{\sqrt{-a}}\right )}{\sqrt{-a}} + 16 \,{\left (b x^{2} + a\right )}^{\frac{3}{2}} + 192 \, \sqrt{b x^{2} + a} a - \frac{165 \,{\left (b x^{2} + a\right )}^{\frac{5}{2}} a^{2} - 280 \,{\left (b x^{2} + a\right )}^{\frac{3}{2}} a^{3} + 123 \, \sqrt{b x^{2} + a} a^{4}}{b^{3} x^{6}}\right )} b^{3} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x^2+a)^(9/2)/x^7,x, algorithm="giac")

[Out]

1/48*(315*a^2*arctan(sqrt(b*x^2 + a)/sqrt(-a))/sqrt(-a) + 16*(b*x^2 + a)^(3/2) + 192*sqrt(b*x^2 + a)*a - (165*
(b*x^2 + a)^(5/2)*a^2 - 280*(b*x^2 + a)^(3/2)*a^3 + 123*sqrt(b*x^2 + a)*a^4)/(b^3*x^6))*b^3